Question Number 213859 by ajfour last updated on 19/Nov/24
Commented by mr W last updated on 19/Nov/24
$$\mathrm{0}<{AB}<\mathrm{1} \\ $$$${no}\:{maximum}\:{or}\:{minimum}\:{exists}. \\ $$
Commented by ajfour last updated on 19/Nov/24
Commented by ajfour last updated on 19/Nov/24
nice, thank you.
Commented by A5T last updated on 19/Nov/24
$$\left({For}\:{the}\:{second}\:{diagram}\right) \\ $$$${Let}\:{circle}\:{with}\:{centre}\:{A},{B}\:\:{have}\:{radius}\:{a},{b}\:{resp}. \\ $$$$\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} }+{b}+{a}=\mathrm{1}\Rightarrow\sqrt{{a}\left({a}+\mathrm{2}{b}\right)}=\mathrm{1}−{a}−{b} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +\mathrm{2}{ab}=\mathrm{1}+{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{a}−\mathrm{2}{b}+\mathrm{2}{ab} \\ $$$$\Rightarrow\mathrm{1}+{b}^{\mathrm{2}} −\mathrm{2}{a}−\mathrm{2}{b}=\mathrm{0}\Rightarrow{a}+{b}=\frac{\mathrm{1}+{b}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\left(\mathrm{1}−{b}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} +{b}^{\mathrm{2}} \Rightarrow{b}^{\mathrm{2}} +\mathrm{2}{b}−\mathrm{1}=\mathrm{0}\Rightarrow{b}=−\mathrm{1}+\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{AB}={a}+{b}=\mathrm{2}−\sqrt{\mathrm{2}} \\ $$
Commented by ajfour last updated on 19/Nov/24
$${Find}\:{maximum}\:{AB}. \\ $$